How to determine whether this improper integral is convergent or not?

Question

The improper integral is

$$\int^1_0\frac{e^x-1-x}{x^2\sqrt{x}}dx.$$

It is from an old calculus exam. It should be convergent. I've looked around on how to solve it but I've had no success at all.

Edit: $0\leq\frac{e^x-1-x}{x^2\sqrt{x}}\leq\frac{e^x}{x^3}$ when $0\leq x\leq 1$ and since $\lim_{x\to0^+}x^3\frac{e^x}{x^3}=1$ the improper integral is convergent. Is this right?

Answer 1

Since $\frac{e^x-1-x}{x^2}\rightarrow\frac{1}{2}$ as $x\rightarrow 0^+$ we can find $\delta>0$ so that $\Big|\frac{e^x-x-1}{x^2}-\frac{1}{2}\Big|<\frac{1}{4}$ for any $x\in (0,\delta)$.

But then $$0<\frac{1}{4\sqrt{x}}< \frac{e^x-x-1}{x^2}\frac{1}{\sqrt{x}}<\frac{3}{4\sqrt{x}}$$ on the domain $x\in (0,\delta)$.

By direct comparison $\int_0^\delta \frac{e^x-x-1}{x^2\sqrt{x}}dx$ converges and hence so must your integral.

Answer 2

Near $x=0$, using the expansion of the exponential function you have: $$\frac{e^x-1-x}{x^2\sqrt x }\sim \frac{x^2/2}{x^2\sqrt x}=\frac12\frac1{\sqrt x} $$ Can you take it from here?